Sunday, October 16, 2011

More practice solving trig equations.

Find the values of x between 0 and 2pi where these equations are true, written in radians as multiples of pi.

a) tanx - 6 = 0
b) sinx - ½tanx = 0
c) cos²x = ½

Answers in the comments.

1 comment:

  1. a) tanx - 6 = 0

    This is the same as

    tanx = 6, so we take inverse tangent of 6 and get 1.405647... If we divide by pi, the value is about .4474pi. With tangent, the second angle where this works is pi away, which means

    x = 0.4474pi and 1.4474pi

    b) sinx - ½tanx = 0

    Rewriting, we get

    sinx - ½sinx/cosx = 0

    Factoring out sine we get

    sinx(1 - ½*1/cosx) = 0

    This means either sinx = 0 or

    1 = ½*1/cosx, which solves to

    ½ = cosx

    Sine is 0 at 0, pi and 2pi. Cosine is ½ at pi/3 and 2pi/3

    c) cos²x = ½

    Take square roots of both sides and we get

    cosx = +/- sqrt(2)/2

    This is true at four "famous" angles,
    pi/4, 3pi/4, 5pi/4 and 7pi/4.

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